Question:

The equation of the plane which is parallel to \(x+3y+5z+1=0\) and \(5\) units from the origin is

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A plane parallel to \(ax+by+cz+d=0\) has the same \(a,b,c\), and distance from origin is \(\frac{|d|}{\sqrt{a^2+b^2+c^2}}\).
  • \(x+3y+5z+5=0\)
  • \(x+3y+5z+5\sqrt{35}=0\)
  • \(x+3y+5z-5\sqrt{35}=0\)
  • \(x+3y+5z-3\sqrt{35}=0\)
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The Correct Option is B

Solution and Explanation

Concept:
Two planes are parallel if their normal vectors are parallel. The given plane is \[ x+3y+5z+1=0 \] So any plane parallel to it will have the form \[ x+3y+5z+d=0 \] where \(d\) is a constant.

Step 1: Use the distance formula.
The distance of the plane \[ ax+by+cz+d=0 \] from the origin \((0,0,0)\) is \[ \frac{|d|}{\sqrt{a^2+b^2+c^2}} \] Here, \[ a=1,\quad b=3,\quad c=5 \] Therefore, \[ \sqrt{a^2+b^2+c^2} = \sqrt{1^2+3^2+5^2} \] \[ =\sqrt{1+9+25} \] \[ =\sqrt{35} \]

Step 2: Given distance is \(5\).
\[ \frac{|d|}{\sqrt{35}}=5 \] \[ |d|=5\sqrt{35} \] Therefore, \[ d=\pm 5\sqrt{35} \]

Step 3: Select the answer according to the given options.
The option marked in the paper is \[ d=5\sqrt{35} \] Hence the required plane is \[ x+3y+5z+5\sqrt{35}=0 \]

Step 4: Final answer.
\[ \boxed{x+3y+5z+5\sqrt{35}=0} \]
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